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The curve f(x)=(x^2+a x+b)/(x-10) has a ...

The curve `f(x)=(x^2+a x+b)/(x-10)` has a stationary point at `(4,1)` . Find the values of `a a n d b` . Also, show that `f(x)` has point of maxima at this point.

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The correct Answer is:
a=7,b=-6

f(x) =`(x^(2)+ax+b)/(x-10)` has a stationary point at(4,1) so , it must lie on the curve this
16+4a+b=-6
Also `(dy)/(dx)=0`
or `(x^(2)-20x-10x-b)/(x-10^(2))=0`
or 10a+b=-64
from 1 and 2 we have a =-7,b =6
also for these values of x ,
`y=(x^(2)-7x+6)/(x-10)`
`therefore (dy)/(dx)=((x-4)(x-16)/(x-10)^(2))`
Hence x=4 is point of maxima
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